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// A set iterator can cope with removing the current entry.
function test(letters, toRemove) {
var set = new Set(letters);
toRemove = new Set(toRemove);
var leftovers = [...set].filter(x => !toRemove.has(x)).join("");
var log = "";
for (let x of set) {
log += x;
if (toRemove.has(x))
set.delete(x);
}
assertEq(log, letters);
var remaining = [...set].join("");
assertEq(remaining, leftovers);
}
test('a', 'a'); // removing the only entry
test('abc', 'a'); // removing the first entry
test('abc', 'b'); // removing a middle entry
test('abc', 'c'); // removing the last entry
test('abc', 'abc') // removing all entries